Is (x+y)(x-y) = even integer?

Question

Is  (x+y)(x-y) =  even integer?

1)  x^2+y^2 =  even

2)  x+y = even

Source: self-made

KarishmaB · Accepted Answer

So now the question is - what is the actual answer?

Try using both statements together: If you know that  x + y  is an even integer and  x^2 + y^2  is an even integer, is it necessary that  x^2 - y^2  should be an even integer too? Actually, no!

It is easy to show that if x and y are both even/both odd integers, then  x+y  is even,  x^2 + y^2  is even and  x^2 - y^2  is even.

But how do you figure out a case where  x+y  is even,  x^2 + y^2  is even but  x^2 - y^2  is not an even integer?

Let me give you an example: Say  x = 3+\sqrt{6}  and  y = 3 - \sqrt{6} .

Here,  x + y = 6  (even integer)
 x^2 + y^2 = 30  (even integer)
But  x^2 - y^2 = 12\sqrt{6}

So answer is (E).

What you should think about is: How can you prove with variables that answer is (E)? Perhaps, the form of x and y that I have given as an example can help you!

Bowtye · Answer

Am i missing something simple?

A)(x+y)(x-y) = x^2+y^2 and we are told that is even... so suff

B) we are told (x+y) is even. even * even = even and odd * even = even . Even if (x-y) = 0, gmat considers 0 as even. So Suff

Harley1980 · Answer

about A)  (x+y)(x-y)  not equal to  x^2+y^2  it is equal to  x^2-y^2

about B)
Let's assume that  x =\frac{5}{3}  and  y = \frac{1}{3} ,  x + y = \frac{5}{3}+\frac{1}{3} = \frac{6}{3} = 2

but  x^2+y^2  will be equal to  (\frac{5}{3})^2 + (\frac{1}{3})^2 = \frac{25}{9}+\frac{1}{9} =\frac{26}{9}  not even

KarishmaB · Answer

Using the same logic, think what happens in stmnt 1 when  x^2 = 5/3  and  y^2 = 1/3 .
In this case,  x^2 + y^2 = 2  (even) but  x^2 - y^2 = 4/3  (not an even integer)
but if  x^2 = 12  and  y^2 = 6 , both  x^2 + y^2  and  x^2 - y^2  are even. 
So how can the answer be (A)?

Harley1980 · Answer

Wow, that's amazing.  VeritasPrepKarishma , you have an eagle eye )
Thank you.

Zhenek · Answer

My solution:
For the first option:  X^2 + Y^2 =  even, then if you add or subtract  2*X*Y , you will get an even number aswell, so  X^2 + Y^2 + 2*X*Y = (X + Y)^2  is even and  (X - Y)^2  is even too. Say both of them are  2*k  and  2*b  respectively.
 \sqrt{2*k}*\sqrt{2*b} = 2*\sqrt{k}*\sqrt{b}  which is even, thus #1 is sufficient.
For the second - well, its quite obvious: product of even number with whatever = even, thats true, thus sufficient

C it is then 
edit: meant D, which is, sadly, incorrect.

Harley1980 · Answer

Zhenek , this is look as you say that answer D is right: both statements are sufficient by themselves. Am I right?

Zhenek · Answer

Oh, yea, I guess I meant that indeed, not experienced with the gmat thing yet. Looks like my answer is wrong then, what a bummer. I guess I really need to pay attention to the given info (no information given about X and Y being non-integers, which completely blows my solution from the get-go:  2*X*Y  could be  2*\sqrt{5}/2*\sqrt{3}/2  which is not even remotely even )

Zhenek · Answer

Well then, lets amend my approach. Lets start from get-go.

Looking at 1st option and taking into account the fact that X and Y could be non-integers, we can't really say anything about the answer to the question with just option 1 on its own, I'd just call it insufficient right away after comming up with 2 different examples, one being integer and another - non-integer.
Easiest ones that come into mind are: 
1)X = 1, Y = 1: 2 is even, 0 is even
2)X =  \sqrt{5}/2 , Y =  \sqrt{3}/2 : 2 is even, 1/2 is not even
This makes #1 insufficient on its own.

#2 - same story, take fractions as second example and integers as first example, insufficient on its own.

Now lets look at them together
 x^2 + y^2 = 2*k 
 x + y = 2*m  =>  y = 2*m - x 
 x^2 + y^2 = (x+y)^2 - 2*x*y = (x+y)^2 - 2*x*(2*m-x) = 4*m^2 -2*x*(2*m - x) = 2*k 
 2*k = 4*m^2 - 2*x*(2*m - x)  =>  x^2 -2*m*x + 2*m^2 - k = 0 
 x = m ± \sqrt{k - m^2} 
 y = 2*m - x = m ∓ \sqrt{k - m^2} 
So we found values of X and Y that would match 2 of these equation(both #1 and #2). That being said, k and m are random positive integers thus we can't be sure if the expression under the root is a perfect square or not.
If you input these values into your question, you will get  x^2 - y^2 = (x-y)*(x+y) = ± 4*m*\sqrt{k - m^2}  which unfortunatelly doesn't answer our question coz of root's value being uncertain (perfect square, then answer is "even", if it is not perfect square, then answer is "not even")
The answer  E , yet again, contradicts the OA, which is unfortunate.

Harley1980 · Answer

Looks like my first attempt to make a task became a train wreck

VeritasPrepKarishma , Zhenek thanks for your explanations, it is now clear. I've changed answer to E. Looks like this isn't 600-700 lvl task.

Hadrienlbb · Answer

I had a hard time with this question.

I would like to better understand what went wrong in my approach:

The question is: \((X + Y)(X - Y)\) = Even?
Which I rephrased too: \(X^2 + Y^2 - 2XY\) = Even?
Hence, when looking at statement 1 and 2, I'm able to asses that X and Y need to either both be Even or Odd, and \(2XY\) has to be Even. So I just went on to select D.

Please help me understand why I can't take this shortcut and I fell right into it.

Thanks,

Please help me understand why I can't take this shortcut and I fell right into it. Thanks,

KarishmaB · Answer

Note that  (X + Y)*(X - Y) = X^2 - Y^2

Hadrienlbb · Answer

OMG

I swear I have not been drinking. Just working too hard! Nonetheless, this is unacceptable of me. Thanks very much Karishma, and forgive my carelessness.

KarishmaB · Answer

It is better to make these errors during practice so that you do not make them during the actual exam!!

mvgkannan · Answer

I think the trap is that it is asked whether (x+y)(x-y) is an even integer, but didn't define that x and y should be an integer. That association is made by us, incorrectly. Nice one.

Is (x+y)(x-y) = even integer?

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